Static and Dynamic
Chapter 1 : Introduction
Introduction to static
•
Mechanics can be defined as that branch of
the physical sciences concern with the state of
rest or motion of bodies that are subjected to
the action forces.
•
Basic mechanics is composed of two principal
areas:
• Static
•
•
Deal with the equilibrium of bodies, that is, those
that are either at rest or move with a constant
velocity
Dynamic
•
Concern with the accelerated motion of bodies.
Fundamental concept
•
Basic terms
•
Length
needed to locate the position of a point in space and
thereby describe the size of a physical system.
• once a standard unit of length is defined, one can then
quantitatively define distances and geometric properties
of a body as multiples of the unit length.
•
•
Space
the geometry region occupied by bodies whose positions
are described by linear and angular measurement
relative to a coordinate system.
• for three-dimensional problems three independent
coordinates are needed.
• for two-dimensional problems only two coordinates will
required.
•
•
Time
the measure of the succession of event and is a basic
quantity in dynamics for three-dimensional problems
three independent coordinates are needed.
• not directly involved in the analysis of static problems
•
•
Mass
a measure of the inertia of a body, which is its
resistance to a change of velocity.
• can be regarded as the quantity of matter in a body.
• the property of every body by which it experiences
mutual attraction to other bodies.
•
•
Force
the action of one body on another.
• tends to move a body in the direction of its action.
• the action of a force is characterized by its magnitude,
by the direction of its action, and by its points of
application.
•
•
Particle
has a mass, but a size that can be neglected.
• Example: the size of the earth is significant compared to
the size of its orbit, therefore the earth can be modeled
as a particle when studying its orbital motion.
• when the is idealized as a particle, the principles of
mechanics reduce to a rather simplified form since the
geometry of the body will not involved in the analysis of
the problem.
•
•
Rigid body
can be considered as a combination of a large number
of particles in which all the particles remain at a fixed
distance from one another both before and after
applying a load.
• as the result, the material properties of any that is
assumed to be rigid will not have to considered when
analyzing the forces acting on the body.
• in most cases the actual deformation occurring in
structures, machines, mechanisms, and the like are
relatively small, and the rigid-body assumption is
suitable for analysis.
•
•
Conversion factors
Newton’s three laws of motion
•
First law
•
A body at rest will remain at rest, and a body in motion
will remain at a uniform speed in a straight line, unless it
is acted on by an imbalanced force.
F1
F2
v
F3
•
Second Law
•
•
A particle acted upon by an unbalanced force, F
experiences acceleration, a that has the same
direction as the force and magnitude that is
proportional to the force
If F is applied t a particle of mass, m, this law may
be expressed mathematically as
F = ma
F
a
Accelerated motion
•
Third Law
•
For every action, there is an equal but opposite reaction..
force of A on B
F
R
force of B on A
Action - Reaction
•
Which person in this ring will be harder to move? The
sumo wrestler or the little boy?
Newton’s law of gravitational
attraction
•
Gravitational attraction between any two particles is gover after formulating
Law of motion
m1m2
F G 2
r
•
Where
F
= force of gravitation between the two particles
G
= universal constant of gravitation; according to
experimental evidence,
m1,m2= mass of each of the two particle
r
= distance between the two particles
Weight
•
What is the different between Mass and Weight?
•
The relationship between mass and weight can be
expressed
•

develop an approximate expression for finding the weight, W
of a particle having a mass m1 = m
Assume the earth to be a non-rotating sphere of constant
density and having a mass m2 = Me, then if r is the distance
between the earth’s center and the particle, we have
W G
Letting,g 
G Me
r2
g = 9.807 m/s2
so
W  mg
m1 M e
r2
yields
Units of measurement
•
Mechanic deal with four fundamental quantities
•
Length
•
Mass
•
Force
•
Time
SI Units in Two system
U.S Customary Units
Units and symbols
Quantity
Dimensional
Symbol
Unit
Symbol
Unit
Symbol
Mass
M
kilogram
kg
slug
-
Length
L
meter
m
foot
ft
Time
T
second
s
second
sec
Force
F
newton
N
pound
lb
•
SI units
•
International system of units
•
Newton (N)
•
Force in Newtons(N) is derived from F=ma
1kg
Force?
•
Solution
N  kg
m
s2
W  mg
(g=9.81m/s2)
•
US Customary
•
The unit of mass, called a slug, is derived from F = ma.
•
Newton (N)
•
Force in Newtons(N) is derived from F=ma
1slug
•
Solution
mass?
lb. sec 2
slug 
ft
w
m
g
(g=32.2 ft/sec2)
Conversion factors
Terms
U.S Customary
S.I metric unit
Length
1 in.
1 ft
1 mile
= 25.4 mm
= 0.3048 m
= 1609 m
Area
1 in.2
1 ft2
1 sq mile
= 6.45 cm2
= 0.093 m2
= 2.59 km2
Volume
1 in3
1 ft3
= 16.39 cm3
= 0.0283 m3
Capacity
1 qt
1 gal
= 1.136 I
= 4.546 I
Mass
1 Ib
1 slug
= 0.454 kg
= 14.6 kg
Velocity
1 in/sec
1 ft/min
I mph
= 0.0254 m/s
= 0.3048 m/s
= 0.447 m/s = 1.61 km/h
Acceleration
1 in./sec2
1 ft/sec2
=0.0254 m/s2
= 0.3048 m/s2
Force
1 Ib
1 poundal
= 4.448 N
= 0.138 N
Pressure
1 Ib/in.2
1 Ib/ft2
= 6.895 kPa
= 47.88 kPa
Energy
1 ft-Ib
1 Btu
1 hp-hr
1 watt-hr
= 1.356 J
= 1.055 kJ
= 2.685 MJ
= 3.6 kJ
Power
1 hp
0.746 kW
Example 1.1
•
Convert 2 km/h to m/s and ft/s
Solution
Since 1 km = 1000 m and 1 h = 3600 s, the factors of
conversion are arranged in the following order, so that
a cancellation of the units can be applied:
km 2 km  1000 m  1 h 
2




h
h  km  3600 s 
m 3.281 ft
 0.556 x
 1.824 ft / s
s
m
Mathematic required
•
Algebraic equations with one unknown
•
Simultaneous equations with two unknowns
•
Quadratic equations
•
Trigonometry functions of a right – angle triangle
•
Sine law and cosine law as applied to non-right angle
triangles.
•
Geometry
•
Algebraic equations with one unknown
Example 1.2
Solve for x in the equation
3(6  x )
 16
2
x?
4
•
Simultaneous equation
Example 1.3
Solve the simultaneous equations.
3x  4 y  8
6 x  2 y  10
x?
y?
•
Quadratic equations
Example 1.4
Solve for x in equation
3x ( 4  2 x )  10  x 2  8
 b  b 2  4ac
use x 
2a
•
Trigonometry functions of a right – angle triangle
r
Sin  
y side opposite

r
hypotenuse
cos  
x side adjacent

r
hypotenuse
tan  
y side opposite

x side adjacent
y

x
•
Sine law and cosine law as applied to non-right angle triangles
•
Triangles that are not right – angle triangles
b
A
C
A
B
C


sin a sin b sin g
g
a
B
•
Side divided by the sine of the angle opposite the side
a
C
b
g
C 2  A2  B 2  2 AB cos g
B
A
•
Right – angle triangle where g = 90o
a
C
b
A
g
B
C 2  A2  B 2
•
Geometry
•
opposite angles are equal when two straight lines
intersect
a
a=b
c=d
d
c
•
b
supplementary angles total 1800
a
b
a + b = 1800
•
complementary angles total 900
a
a + b = 900
b
•
a straight line intersection two parallel lines produces the
following equal angles:
c
d
a
b
a=b
c=d
or
a=b=c=d
•
the sum of the interior angles of any triangles equals to 180o
a
a+b+c=
1800
b
c
•
similar triangles have the same shape
D
A

BD DE BE


BA AC BC

B
C
•
E
If AB = 4, AC = 6 and DB = 10, then by proportion
DE 
6
 10  15
4
•
circle equations:
circumfere nce  D or 2r
Area 
•
D 2
4
or r 2
Angle  is defined as one radian when a length of 1 radius
is measured on the circumference.